Strength of Materials for FEA: Stress, Strain and Principal Stresses

Hello, everyone. Welcome you all in week one, lecture one. So let's have an overview of the course. So in this course, we are going to study about the problems in structural engineering. So here we have divided the entire course into two parts. The first part is the theoretical part in which we are going to learn about the strength of materials and the finite element method. And in the other part, we are going to learn about the software tool. So we are going to learn about the Ansys Workbench and we are going to learn different kinds of analysis which has been performed in the industry.

So this is a course for the structural engineering domain. So a prerequisite for this course is a person who has a degree or a diploma in the mechanical engineering can avail this course and learn the contents which has been described in the course. Okay. So let's start with the first part of our course. That is the strength of material FEA perspective. And this we have the table of contents is this and we are going to cover all this points which you might have studied in your college.

Okay. So this concept is very much useful in the FEA perspective for the structural engineering domain and this concept we are going to use in a day-to-day life when we are going to solve the industry problem using the software. This is the learning objective. So this is the learning objective. So after going through this presentation, the strength will be able to understand the following concepts. This concepts are mentioned here. So let's get started to the introduction of strength of material. Okay. So in the terms of structural engineering, the strength of material is a branch of applied mechanics that deals with the behavior of solid bodies subjected to various loading.

Okay. So what do you understand by this sentence? So the strength of material, it means like by this three words, strength of material. So a person can think of that it is talking about the material system. Okay. Suppose, let's take an example. Okay. So this is a rope. Okay. And there is a tug of war competition. A person is pulling the rope this side. Okay. And another person is also pulling the rope towards his side. Okay. So what do you think about this example? We are just checking the strength of the person as well as the strength of the rope also, right?

So here we have the three materials. Two of them are the person and one is the rope. So if the rope will fail to sustain the load, either from the person one or person two, okay, this is person one and this is person two. So that the rope will fail eventually. But if the rope is too strong to sustain the loading provided by the any of the person, then what happened? It means that the person one, suppose the person one have the more strength than person two, then the person one will able to pull the rope towards his side and win the tug of war competition.

So this is basically the strength of material. The objective of our analysis is to determine the strength of the material. Okay. And how we are going to determine the strength of the material based on what the parameters. So those are the mathematical parameters that is termed as the stress, strain, deflection. Okay. Okay. So these are basically three parameters. What we are doing on the application of the load, we want to know what is happening in the material. Okay. To know what is happening in the material, we have to quantify the data.

Okay. So only the maths can do the quantification of the data. So those quantification of the data is termed as the stress, strain, deflection, force. Okay. Basically, these are the parameters which will determine to study our analysis. Okay. So these are the objective of our study. So stress and strain. So what is the first thing we will strike when you will term as stress or strain? Okay. Because you all know this definition of these two words or these two terminologies. So let's take the normal stress and with the help of the normal stress, because stress can be shear stress, normal stress, inclined stress.

So let's take an example of the normal stress and learn the concept of the stress. Okay. So what is normal stress basically? So let's consider a prismatic bar. So this is the prismatic bar under the external load of P. So what will happen to this bar? A prismatic bar means it is having a uniform cross section throughout its length. Okay. So when we are applying the load P, what this bar actually will do? It will try to resist those external loads, right? Okay. So those external loads can be resisted by this sections.

Okay. Suppose we'll take a section MN and we want to see what is happening basically at the section MN. So at the section MN, suppose we have a unit cross section area A. Okay. So what this unit cross section area A is doing, it is similarly like a tug of war competition. Okay. So what the each particle of this or the each element of this unit area will try to resist this external load P. So this is the internal resistance by the external loading. Okay. And it will come as an intensity of the load.

So why it is called as the intensity of the load? Because it has been distributed over the cross section of the bar. Okay. So this internal resistance per unit area is termed as stress. So this is basically sigma is equals to R by A or it is the static case. So if you'll do the force balance, then the summation of R will be equals to the summation of external load. Okay. So R is equals to P. Okay. So R by A should be equals to P by A.

Okay. Okay. So this is the internal resistance per unit area. So if anyone will ask in the interview, what is the stress? Never ever tell that it is the force per unit area. Okay. Because the force per unit area is the pressure and the pressure is generally the external loading. But we know the stress that it is the internal resistance per unit area. So that is the concept of stress. Hope it is clear for you. So now coming to the strain. So taking the same example, which we have in the previous one to explain the stress.

So let's take the same prismatic bar of uniform cross section of unit area A and we are applying a load P. So when you are applying the load P and what will happen once this load P will apply the load P there, it will start deforming also. Okay. It will start deforming and this deformation will result in the change in the length. So this change in the length DL per unit, the original length is termed as normal strain. Again, I hope it is clear. So it is trying to resist that load.

Okay. Initially, it is trying to resist that load. So in that scenario, the stress has occurred. Now it is trying to like the external load is make this bar to elongate also because of the external load. The bar is getting elongated. Okay. So in that scenario, the concept of strain appear and it termed as the change in length per unit original length. So these are the types of stresses, normal stress, shear stress, and incline stress. The incline stress will have the both component, the normal and the shear.

And the shear stress, if you can see here, it is parallel to the cross section area. So now consider a hammer placed on the table. Will there be any stress on the hammer if we lift it gently? Okay. Ignore the self-weight as of now. So there will be no stress. Obviously, there will be no stress as there is no external force applied. Okay. This phenomena is termed as the rigid body motion. Okay. What is rigid body motion? It means without any physical deformation, when we are displacing the object from one position to the another, then this is termed as a rigid body motion.

So suppose you have XYZ axis, you are moving it in X direction, you are moving it in Y direction, you are moving freely in Z direction. So in all the directions, it is just moving freely. So there is no stress because there is no physical. The physical deformation has been seen in the body. Whenever there will be the physical deformation, it means that it is going under certain stress. So this is for the interview perspective like blue-low carbon steel is having less than 0.24% of the carbon content.

High carbon is still 0.6 to 1.25, medium is 0.25 to 0.6. Okay. As the carbon content increases, the steel will become less ductile. So this is just for the interview purpose. You should know this thing. So now we are going to study about the stress-strain curve. So because of the stress, we have the strain. So it means there's some relationship between the stress and the strain, right? Okay. So let's have a look on the stress-strain curve. So this is the stress-strain curve of the structural steel under the tensile load.

And this is not to scale image. You can see it. There's a little bit exaggerated image. Okay. So if you are referring to some of the standard textbook, then you can see this image. Otherwise, in your textbook, you will have this B to C curve is much close to each other. Okay. So what is happening here? Okay. So we have a UTM machine under which this specimen has been placed and tensile load has been applied like this. So in this scenario, or we can place it horizontally also.

So in this scenario, what will happen? Okay. A term called gauge length in which we are focusing to calculate the stress and strain. So we'll take a gauge length based on the ASTM standard. ASTM is nothing but the American Society for Testing and Materials. So based on the ASTM standards, we can take the gauge length. Okay. And suppose between this point and this point, we can take a gauge length of L like this. Okay. So in this scenario, when we have loaded this prismatic bar under the tensile load, we have seen that the O2A stress is proportional to the strain.

So in this scenario, in this case, what is happening basically that the load is directly proportional to the deflection. Okay. It means that the stress is also directly proportional to the strain. Okay. So it is proportional to the proportionality limit. Okay. So if you can see it, and this is a proportionality limit here. Okay. This and it is very close to the yield stress also. Okay. So what will happen after the point A is the behavior is going to be non-linear. A non-linear behavior starts emerging.

Okay. Okay. So in this scenario, we can say like this, with the increase in the load beyond the proportional limit, the strain begins to increase more rapidly for each increment in this stress. So if you can see from the point B to point C, okay, without any significant change in the load, we have a very considerable strain. And this small part is called as a perfectly plastic or a yielding stage. So point B is the yield point. Okay. And from point B to point C is the perfectly plastic or this zone is called the yielding zone.

Beyond point C, what will happen? Some disturbance or something happens at the molecular level, at the atomic level, which will change the crystalline structure of the metal and the strain hardening will occur. Okay. So in this scenario, from point C to D, it will go under the strain hardening zone. It will go into the strain hardening process because the material undergoes a change in its atomic and crystalline structure. Okay. So once the strain hardening will happen, because of the strain hardening, you will have to put extra effort to get more deflection or more strain.

Okay. So in that scenario, you have to increase your stress and to get the considerable amount of strain in our specimen. Okay. So in this, we have a nonlinear zone like this here, C to D and sigma is equals to K epsilon to the power N. Okay. So this is a kind of parabolic equation and this sigma is equals to K epsilon to the power N. N is the strain hardening coefficient. Okay. So what will happen now is there is a lateral contraction of the bar will happen.

Okay. And what we are drawing is as of now is the engineering stress strain curve. Okay. So in the engineering stress strain curve, whatever we are calculating is based on the original cross-section area. Okay. Try to understand this concept. Till point C, the lateral contraction of this bar is not too much considerable. After the point C, the lateral contraction will start showing its own face. Okay. Now it will play a major role in the stress strain curve. Okay. Now point D is the ultimate strength in the engineering stress strain curve.

Around the proximity of the D, the necking will start to take place. Okay. When the necking will start to take place, the force required to reload or the force required to stretch further will reduce. And the fracture finally will happen at point E. Okay. So there is a considerable reduction of the load has been observed and with the decrease in the cross-section area. And thus the fracture will happen at the point D. The area is changing continuously. And it is based on the previous area. Like, like, suppose at this point of time.

Yeah. At this point, at this, we have area A0 at 0. And now this is area AD. Now the cross-section area AD is lesser than the cross-section area A0 or AO. Okay. So the point E dash. Okay. So O, A, B, C, D, E is the engineering stress strain curve. And O, A, B, C, D, E dash is true stress strain curve. Okay. So this is the concept of the stress strain curve, which we have understood with the help of the structural strain or the mild strain problem.

Okay. Hope it is clear. So linear elasticity, Hooke's law, and Poisson's ratio. In the previous stress strain curve, we have observed most structural material have an initial stress strain curve in which material behaves elastically and linearly. Okay. So this is a very important line with respect to FEA. FEA perspective, this sentence is, has a lot to say. Okay. So let's go to this stress strain curve and re-look what does it means. Here what it is telling, most structural material have an initial region on stress strain diagram in which material behaves elastically and linearly.

Okay. So what is linear? Linear means Y equals to MX plus C. A straight line equation. Okay. Y equals to MX plus C. Y equals to MX plus C. So Y equals to MX plus C is there. Okay. So here Y is sigma is equal to MX plus C. Y equals to MX plus C. Okay. Now, as of now, you remember this equation, okay, to study the linear zone. Okay. So sigma is equal to MX plus C equals to MX plus M. So from point O to point A, the stress is directly proportional to the strain.

And when the stress is directly proportional to the strain, and when you remove the proportionality, you will have to add a constant. And it is going to be sigma is equal to Epsilon E. Okay. So what hooks say for a matter? which is homogeneous and isotropic material. There is a mathematical relationship between the stress and the strain, okay, and the stress and the strain is directly proportional to the each other and the constant of proportionality is called as the modulus of elasticity which is nothing but the slope of the stress strain curve in the linear elastic region.

So now you got the term of the linear. What is linear? The linear is nothing but the straight line, equation of the straight line. This is the equation of the straight line, sigma is equals to, so now M is equals to E, okay, sigma is equals to E epsilon, okay. So this is the equation of the straight line, okay. Then what is the meaning of the elastically? When material behave elastically and linearly. So one concept we have understood is linear material. The second concept is elastic. So what is elastic material?

So you try example of the rubber, okay. So you try example of the rubber, okay. The rubber when you stretch it to a limit, it will come to its original position. Okay, that is the meaning of the elasticity. Okay, so what is happening here with the steel, when we have loaded the bar or the specimen till the proportional limit and then we have unloaded it, there is fully regain of its shape and size, okay. So when we have loaded our material with some value of stress, then it has a strain value. That is the linear region. So it will have, suppose this is Epsilon, okay.

So when we'll unload it, Epsilon will turn to zero and there will be no residual strain has been seen or has been observed after unloading of the material when the material is loaded till the proportional limit, okay. So that is the elastic behavior of the material, okay. So this is a very important concept. The most of the FEA problems we are going to solve by assuming the material is linear and elastic, okay. So keep this in mind. Okay, so this is the same. Taking the example of the structure is steel from or isn't to proportional.

There is a linear relationship between the stress and the strain and material behaves elastically. So material is said to be linear elastic from O to A. It is the same what I have explained, okay. So this is the Hooke's law, which we have studied and just a few points on the modulus of elasticity. It is the slope of the stress strain curve, okay. And E is the property of the material, okay. So now observation two. So here we have two prismatic bars, one of steel and another of aluminum.

Both are of same dimensions and static external force is acting on it is also along the same vector as shown in the figure. In which material the stress is going to be the higher. Also the Young's modulus of the steel is 210 GPA and aluminum is 70 GPA. So what is the definition of the stress? So the stress is defined as the internal resistance per unit area. So by that definition, are you observing anything like depending on the steel or the aluminum's Young's modulus? No, right.

So in that scenario, the stress at any section of the bar will be same for both the materials for the given static load because the stress is a function of a external load resisting cross-section area. Okay. And stress is not a function of material. So this is a very important question is going to clear your concept of our distress, strain, Young's modulus. Okay. How are these three entities are related or not related to each other? Okay. Hope this is clear to you. So what is Poisson's ratio?

When a prismatic bar is loaded under the tension due to the axial elongation, it is also accompanied by lateral contraction. Okay. If you see here, this red one is the original dimension of the prismatic bar. We have applied the load. It has elongated by this length and it has been contracted by this length. So we also know that the isotropic material have the same elastic property in all the directions. So this E is delta V divided by original volume. It should be delta V divided by original volume.

So E is basically called dilation. So this equation will, the change in volume per unit volume will result to this equation, sigma by epsilon multiplied by 1 minus 2 nu. Okay. So what is this equation is doing here is basically it is limiting the value of the Poisson's ratio. Because this change in volume divided by the original volume, it is not going to be the negative. Okay. It has to be positive in that scenario and E is called the dilation. So as the change in volume per unit volume is always positive and the sigma by E multiplied by 1 minus 2 nu.

If you equate it to 0, that is the limiting condition, you will get the maximum Poisson's ratio. So if you do this math, 1 minus 2 nu should be 0 and 2 nu is equals to 1. So the maximum value of the limiting value or the maximum value of the Poisson's ratio should be 0.5. And the Poisson's ratio should be, okay, should be 0 to 0.5. Okay. And 0.5 is the limiting value and it is for the rubber. Rubber have this limiting value 0.5 and the lower limiting value is for corp.

Corp will have a Poisson's ratio of the 0. So that's why we use it to tighten the bottle. This is all with the Poisson's ratio. This is the material is homogeneous if it has the same composition at every point in the body. If the material is homogeneous and isotropic, the lateral strain is directly proportional to the axial strain. Okay. So this Poisson's ratio is limited to the homogeneous and isotropic material. When the material is homogeneous and isotropic, the lateral strain is directly proportional to the axial strain and the Poisson's ratio is the constant of proportionality.

So the negative symbol here is because the lateral strain is negative. If the axial strain is positive, then the lateral strain is negative. So we have a add-on negative symbol here. Okay. So this is all about the Poisson's ratios. So what is Young's modulus? We have already discussed about it. Here we have discussed about the Young's modulus mathematically. Okay. It is the property of material and it is the slope of the linear elastic region of the stress strain curve. Okay. So can anyone tell me like which material will be more stiffer?

Now we'll have a discussion on the deflection. So when we have the Young's modulus of steel S210 GPA and aluminium of 70 GPA, can you now tell that which material is going to deflect more? First part, we have understood. Now I want to know which material is going to deflect more. So it is obviously the aluminium because it is less stiffer than the steel. Okay. So this is also one of the important observations we have. Now we'll go to the elasticity and plasticity. So here are two curves or two stress strain curves, which is going to explain about the elasticity and plasticity.

Okay. So suppose we have a stress strain curve like this. So from O to E, E is the elastic limit, this C is the elastic limit, this F is the failure point, A or point B is the point up to which we have loaded our material. So if we load the material till point A, okay, if we load the material till point A and then we'll unload it, what will happen? It is going to regain its shape and size completely. Now what will happen if we load our material till point B, which is crossing the elastic zone?

Now we have the elastic zone till here, okay, till here. Now if we have loaded this, the material beyond point T, what is going to happen? It went to the inelastic zone. So from E to F, it is called as basically inelastic or plastic zone. Okay. Now. What will happen? When we'll unload this material now from point B. So we have loaded our material till point B, which is in the plastic zone or in elastic zone. And then we have unloaded it. Then there is a residual strain and the elastic recovery. Okay. So it will recover till this limit. Okay.

It is going to recover only this, this much. So C to D will be the elastic recovery and there will be a residual strain. Okay. Now, suppose we have a material after unloading, it came like this. So in that scenario, we have this much as a elastic recovery and this much as a residual strain. Okay. It's because this much part of the strain, it has recovered only this much left. So this is the residual strain or the permanent set in the material. Okay. So this concept is very much important for the explicit problem where we are going to solve the impact test, drop test.

So in those scenarios, in those scenarios, we are going to see how much strain or the permanent set or the residual strain is there on the body after the collision or after the impact with the barrier or like the impact of the car with the barrier like that. Okay. So this is a very key point of solving the problem in the explicit analysis. So those things we are going to learn in the further classes. Okay. So here is the explanation. Suppose a material is loaded statically in tension and compression. Now, if we unload the material slowly, there are two possibilities.

If the loading is within the elastic limit, the material flows exactly the same curve and back to origin and regain its original dimension. If it is loaded beyond the elastic limit, that is inelastic or plastic zone, it will partially regain its size with a residual strain or permanent strain in the material. Now, we have a problem here. Now, we have a problem here. This is a bending moment problem of a cantilever beam. One side fixed and another side is loaded. Now, sigma equals to MYYI and sigma equals to EY by R.

I want to know one thing here. What is the parameter like if we are going to change the moment of inertia or the Y, that is the distance from the centroidal axis to the fiber on which we are going to calculate the stress? So if we are going to change all this three, what is the effect on the stress and what is the effect on the strain? So the question number one is, what if we will change the Young's modulus? Suppose we have the same question here.

This is suppose this is of steel and one more similar, same. Aluminium material we have and same cantilever beam. What will be the bending stress? Will it be same for this both material or it will be different? Because in this equation, in this equation, we have the sigma Y by Y is equals to MYY is equals to E by R. Okay. So if you will calculate it, you will find the stress value. The stress value is not a function of this E. Sigma is equals to MYY.

It is going to come correctly if we don't know the Young's modulus also. Okay. But when it comes to the deflection, we should have a Young's modulus in the equation because the deflection will be more when the stiffness is less and the stiffness is a function of Young's modulus. So try to correlate this with the normal strain condition, which we have just discussed in the observation too. So now coming to the principal stresses. What is principal stresses? So I hope you might have familiarity with this concept also.

Okay. So the first, we'll go through the principal plane. Principal plane are those planes in which the material where the resultant stress across them are only normal stress or the plane across which no shearing stress occur. So it is a family of planes in which the resultant stress are normal and there are no shear stress occur on those planes. So principal stress are those stress which are acting on these principal planes. The plane carrying the maximum normal stress is called major principal plane and the stress acting on it is called the major principal stress.

So let's have a look. Okay. Suppose in the given coordinate system on this inclined plane, this inclined plane, we have, we'll take a small element and we have seen that on this element we have a component of sigma y tau xy sigma x tau xy like this. We have the component of the stress. So what we have done, we have rotated the element at such an angle so that at a particular angle, the shear stress will disappear. The shear stress will become zero. The, at that angle, that is the theta p, there will be a principal plane at which the shear stress will become zero and only the normal stress, that is the principal stress will left acting on this element.

So these are called the principal stress and acting on the principal plane theta p. So this is the Mohr circle. It gives too many on information by just drawing a diagram or a circular diagram of a normal stress and the shear stress. So this is a normal stress, shear stress plot. So here you have to plot like this. Suppose on an element we have a sigma y tau y x sigma x tau xy like this sigma y so we'll take a plane. Okay. We'll split this element in the diagonally half.

Okay. So a, b and c. So on plane a, b we have the sigma x and tau x, tau x, y and on plane a, c we have the sigma y, sigma y and tau y, x tau x, y and tau y, x are same. It is a complementary. So this is on one plane, this is on one plane. So this, this are basically the coordinates which we have to put like this sigma x tau x, y and sigma y minus tau x, y. Okay. So here you have to see one more thing.

Also that is sigma x tau x, y. If you see here this is in anticlockwise direction. Okay. So this anticlockwise direction will take it as a positive. Okay. And we'll put that coordinate in the fourth quadrant. Okay. Like here. Okay. And similarly sigma y and tau y, x it will produce a clockwise effect. And this is built on as a negative. So here also, here also it will be negative and sigma y minus tau x, y. Okay. So like this, we'll draw a line and the center of that line will draw a circle.

And the circle where it cuts on the sigma plane, it is the principal stresses. And this is the maximum shear stress, tau max, tau max, which is nothing but the sigma 2 minus sigma 1 by 2. Okay. So that is basically the radius of the Mohr circle, which will give you the maximum shear stress. Okay. So if you take this plane, A, B, C. So on plane AB, we have the 80 compressive and 25 MPA. So 25 MPA is in clockwise direction. So it is minus 80 and minus 25.

So minus 80 and minus 25 is going to come at this point. And similarly, if you see the 50 and 25, if you see this anticlockwise and this is the 50 tensile. Yeah. So 50 and minus 25 is going to come at this area. You just join it. We'll draw it. And we'll draw a circle with the center of this intersecting on the x-axis. Okay. So this is the Mohr circle we have drawn. So this is the Mohr circle we have drawn.